Explain Vogel’s approximation method by obtaining initial basic feasible solution of the following transportation problem.

Destination | ||||||

D_{1} |
D_{2} |
D_{3} |
D_{4} |
Supply | ||

O_{1} |
2 | 3 | 11 | 7 | 6 | |

Origin | O_{2} |
1 | 0 | 6 | 1 | 1 |

O_{3} |
5 | 8 | 15 | 9 | 10 | |

Demand | 7 | 5 | 3 | 2 |

#### Solution

Total supply (a_{i}) = 6 + 1 + 10 = 17

Total demand (b_{j}) = 7 + 5 + 3 + 2 = 17

`sum"a"_"i" = sum"b"_"j"`.

So given transportation problem is a balanced problem and hence we can find a basic feasible solution by VAM.

**First allocation:**

D_{1} |
D_{2} |
D_{3} |
D_{4} |
(a_{i}) |
penalty | |

O_{1} |
2 | 3 | 11 | 7 | 6 | (1) |

O_{2} |
1 | 0 | 6 | ^{(1)}1 |
1/0 | (1) |

O_{3} |
5 | 8 | 15 | 9 | 10 | (3) |

(b_{j}) |
7 | 5 | 3 | 2/1 | ||

penalty | (1) | (3) | (5) | (6) |

The largest penalty is 6.

So allot min (2, 1) to the cell (O_{2}, D_{4}) which has the least cost in column D_{4}.

**Second allocation:**

D_{1} |
D_{2} |
D_{3} |
D_{4} |
(a_{i}) |
penalty | |

O_{1} |
2 | ^{(5)}3 |
11 | 7 | 6/1 | (1) |

O_{3} |
5 | 8 | 15 | 9 | 10 | (3) |

(b_{j}) |
7 | 5/0 | 3 | 1 | ||

penalty | (3) | (5) | (4) | (2) |

The largest penalty is 5.

So allot min (5, 6) to the cell (O_{1}, D_{2}) which has the least cost in column D_{2}.

**Third allocation:**

D_{1} |
D_{3} |
D_{4} |
(a_{i}) |
penalty | |

O_{1} |
^{(1)}2 |
11 | 7 | 1/0 | (5) |

O_{3} |
5 | 15 | 9 | 10 | (4) |

(b_{j}) |
7/6 | 3 | 1 | ||

penalty | (3) | (4) | (2) |

The largest penalty is 5.

So allot min (7, 1) to the cell (O_{1}, D_{1}) which has the least cost in row O_{1}.

**Fourth allocation:**

D_{1} |
D_{3} |
D_{4} |
(a_{i}) |
penalty | |

O_{3} |
^{(6)}5 |
15 | 9 | 10/4 | (4) |

(b_{j}) |
6/0 | 3 | 1 | ||

penalty | – | – | – |

**Fifth allocation:**

D_{3} |
D_{4} |
(a_{i}) |
penalty | |

O_{3} |
^{(3)}15 |
^{(1)}9 |
4/3/0 | (6) |

(b_{j}) |
3/0 | 1/0 | ||

penalty | – | – |

The largest penalty is 6.

So allot min (1, 4) to the cell (O_{3}, D_{4}) which has the least cost in row O_{3}.

The balance 3 units is allotted to the cell (O_{3}, D_{3}).

We get the final allocation as given below.

D_{1} |
D_{2} |
D_{3} |
D_{4} |
Supply | |

O_{1} |
^{(1)}2 |
^{(5)}3 |
11 | 7 | 6 |

O_{2} |
1 | 0 | 6 | ^{(1)}1 |
1 |

O_{3} |
^{(6)}5 |
8 | ^{(3)}15 |
^{(1)}9 |
10 |

Demand | 7 | 5 | 3 | 2 |

**Transportation schedule:**

O_{1} → D_{1}

O_{1} → D_{2}

O_{2} → D_{4}

O_{3} → D_{1}

O_{3} → D_{3}

O_{3} → D_{4}

i.e x_{11} = 1

x_{12} = 5

x_{24} = 1

x_{31} = 6

x_{33} = 3

x_{34} = 1

Total cost is given by = (1 × 2) + (5 × 3) + (1 × 1) + (6 × 5) + (3 × 15) + (1 × 9)

= 2 + 15 + 1 + 30 + 45 + 9

= 102

Thus the optimal (minimal) cost of the given transportation problem by Vogel’s approximation method is ₹ 102.